The absolute maximum value of y = x3 – 3x + 2, 0 ≤ x ≤ 2, is

i)

$f(x)=x^3−3x+2;0⩽x⩽2$

$f′(x)=3x^2−3=0⇒x=±1$

critical points are +1 & -1, but we have to find absolute maximum in range [0,2] only. Hence we will reject -1.

ii)

Evaluating f(x) at critical points & at end points:

f(1) = 1 − 3 + 2 = 0

f(0) = 0 − 0 + 2 = 2

f(2) = 8 − 6 + 2 = 4

iii)

From above we can see f(x) achieves maximum value of 4 at x = 2 in range x ϵ [0,2]